Codeforces #614 div.2 (A-E)

A  ConneR and the A.R.C. Markland-N

 #include <bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define inc(i, l, r) for (int i = l; i <= r; i++)

 int t, n, s, k, x;

 int main() {
     cin >> t;
     while (t--) {
         cin >> n >> s >> k;
         map<int, int> m;
         inc(i, , k) {
             cin >> x;
             m[x] = ;
         }
         inc(del, , k) {
             if ((s + del <= n && m[s + del] == ) ||
                 (s - del >=  && m[s - del] == )) {
                 cout << del << "\n";
                 break;
             }
         }
     }
 }

B  JOE is on TV!

 #include <bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define inc(i, l, r) for (int i = l; i <= r; i++)

 int n;
 double r;

 int main() {
     cin >> n;
     for (int i = n; i >= ; i--) r += 1.0 / i;
     printf("%.12f", r);
 }

C  NEKO's Maze Game

每个岩浆点的对面或斜对面如果也有岩浆点，那么这两点之间就会形成障碍，统计这种障碍的数量

 #include <bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define inc(i, l, r) for (int i = l; i <= r; i++)

 const int maxn = 1e6 + ;

 int a[][maxn];
 int n, q, k, x, y;

 int main() {
     cin >> n >> q;
     inc(i, , q - ) {
         cin >> x >> y;
         x--;
         if (a[x][y] == ) {
             a[x][y] = ;
             inc(del, -, ) if (a[ - x][y + del] == ) k++;
         } else {
             a[x][y] = ;
             inc(del, -, ) if (a[ - x][y + del] == ) k--;
         }
         if(k) puts("NO");
         else puts("YES");
     }
 }

D  Aroma's Search

分析点集的特点，可以知道每一个数据点都在下标比它小的数据点的右上方，而且离下一个数据点的距离 大于 离第0个数据点的距离。

枚举Aroma从哪个点出发；先往第0个数据点走，然后尽可能往右上走

 #include <bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define inc(i, l, r) for (int i = l; i <= r; i++)

 const int maxn = 1e6 + ;

 ll a0, b0, ax, bx, ay, by;
 ll xs, ys, t;

 ll dx[maxn], dy[maxn];
 ll dis[maxn];

 int res;

 int main() {
     cin >> a0 >> b0 >> ax >> ay >> bx >> by;
     cin >> xs >> ys >> t;
     ll tx = a0, ty = b0;
     int top;
     for (top = ; tx <= 2e16 && ty <= 2e16; top++) {
         dx[top] = tx, dy[top] = ty;
         tx = tx * ax + bx, ty = ty * ay + by;
     }
     inc(i, , top - ) dis[i] = dx[i] - dx[] + dy[i] - dy[];
     inc(i, , top - ) {
         ll tmp = t - abs(xs - dx[i]) - abs(ys - dy[i]);
         if (tmp < ) continue;
         int tot = ;
         if (tmp <= dis[i]) {
             tot = i - (lower_bound(dis, dis + top, dis[i] - tmp) - dis) + ;
         } else {
             tot = upper_bound(dis + i + , dis + top, tmp - dis[i]) - dis;
         }
         res = max(res, tot);
     }
     cout << res;
 }

E  Xenon's Attack on the Gangs

最大S的方案必然是0-L都在一条链上，这条链上的数字是“山谷”型

推出状态转移方程

dp[u][v] = sub[u][v] * sub[v][u] + max(solve(par[v][u], v), solve(par[u][v], u));

 #include <bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define inc(i, l, r) for (int i = l; i <= r; i++)
 #define pb push_back

 const int maxn = 3e3 + ;
 vector<int> edge[maxn];
 int n, u, v;
 int sub[maxn][maxn], par[maxn][maxn];
 ll dp[maxn][maxn], res;

 int root;

 int dfs(int x, int p) {
     sub[root][x] = ;
     for (int i = ; i < edge[x].size(); i++) {
         if (edge[x][i] != p) {
             par[root][edge[x][i]] = x;
             sub[root][x] += dfs(edge[x][i], x);
         }
     }
     return sub[root][x];
 }

 ll solve(int u, int v) {
     if (u == v) return ;
     if (dp[u][v]) return dp[u][v];
     return dp[u][v] = sub[u][v] * sub[v][u] +
                       max(solve(par[v][u], v), solve(par[u][v], u));
 }

 int main() {
     cin >> n;
     inc(i, , n - ) {
         cin >> u >> v;
         edge[u].pb(v);
         edge[v].pb(u);
     }
     inc(i, , n) {
         root = i;
         dfs(root, -);
     }
     inc(i, , n) inc(j, , n) {
         res = max(res, solve(i, j));
     }
     printf("%lld\n", res);
 }